Question

Let $d_1$ and $d_2$ be the lengths of the perpendiculars drawn from any point of the line $7x - 9y + 10 = 0$ upon the lines $3x + 4y = 5$ and $12x + 5y = 7$ respectively. Then

• A. $d_1 > d_2$
• B. $d_1 = d_2$
• C. $d_1 < d_2$
• D. $d_1 = 2d_2$

Answer 1

Answer: (B)

Let $(h, k)$ be any point on the line $7 x-9 y+10=0$,
then $7 h-9 k+10=0$
$\Rightarrow 7 h =9 k-10$
$\Rightarrow h =\frac{9 k-10}{7}$ ... (i)
Now, perpendicular distance from point $(h, k)$
to the line $3 x+4 y=5$ is $d_{1}$
$ d_{1}=\frac{3 h+4 k-5}{\sqrt{3^{2}+4^{2}}} $
$\Rightarrow d_{1}=\frac{3 h+4 k-5}{5}$
$\Rightarrow d_{1}=\frac{3 h+4 k-5}{5}$ ... (ii)
and perpendicular distance from $(h, k)$ to the line $12 x+5 y=7$ is $d_{2}$
$\therefore d_{2}=\frac{12 h+5 k-7}{\sqrt{12^{2}+5^{2}}} $
$\Rightarrow d_{2}=\frac{12 h+5 k-7}{13}$ ... (iii)
Now, $d_{1}-d_{2}=\frac{3 h+4 k-5}{5}-\frac{12 h+5 k-7}{13}$
$\Rightarrow d_{1}-d_{2}$
$=\frac{13(3 h+4 k-5)-5(12 h+5 k-7)}{65}$
$=\frac{39 h+52 k-65-60 h-25 k+35}{65}$
$=\frac{-21 h+27 k-30}{65}$
$=\frac{-21\left(\frac{9 k-10}{7}\right)+27 k-30}{65} $
$=\frac{-27 k+30+27 k-30}{65}=0$ [fromEq.(i)]
$\Rightarrow d_{1}-d_{2}=0 $
$\Rightarrow d_{1}=d_{2}$